What will be the minimum uncertainty in velocity of a particle of mass 1.1×10−28 kg if uncertainty in its position is 3×10−10 cm? (h=6.62×10−34kgm2s−1)
1.6×105ms−1
2.6×105ms−1
6.1×105ms−1
3.6×105ms−1
A
1.6×105ms−1
B
6.1×105ms−1
C
3.6×105ms−1
D
2.6×105ms−1
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Solution
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According uncertainity principle,
Δx.Δp=h/4π
Δx(m.Δv)=h/4π
Δv=(h/4π)×(1/m.Δx)
=6.62×10−344×3.14×1(1.1×10−28)(3×10−12)ms−1
=1.6×105ms−1
Hence, the correct option is A
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